On 10/23/2013 5:53 AM, Bruno Marchal wrote:

On 22 Oct 2013, at 19:01, meekerdb wrote:On 10/22/2013 5:48 AM, Bruno Marchal wrote:On 21 Oct 2013, at 20:07, meekerdb wrote:On 10/20/2013 11:12 PM, Russell Standish wrote:On Sun, Oct 20, 2013 at 08:09:59PM -0700, meekerdb wrote:Consistency is []p & ~[]~p. I was saying []p & ~p, ie mistaken belief.ISTM that Bruno equivocates and [] sometimes means "believes" and sometimes "provable".And I'm doing the same. It's not such an issue - a mathematician will only believe something if e can prove it.But provable(p)==>p and believes(p)=/=>p, so why equivocate on them?Both provable('p') -> p, and believe('p') -> p, when we limit ourself to correct machine.(And incidentally mathematicians believe stuff they can't prove all the time - that'show they choose things to try to prove).Then it is a conjecture. They don't believe rationally in conjecture, when they arecorrect.They believe it in the real operational sense of "believe", they will bet their wholeprofessional lives on it. How long did it take Andrew Wiles to prove Fermat's lasttheorem? Since one can never know that one is a "correct machine" it seems to me amuddling of things to equivocate on "provable" and "believes".On the contrary. It provides a recursive definition of the beliefs, by a rational agentaccepting enough truth to understand how a computer work.We can define the beliefs by presenting PA axioms in that way Classical logic is believed, '0 ≠ s(x)' is believed, 's(x) = s(y) -> x = y' is believed, 'x+0 = x' is believed, 'x+s(y) = s(x+y)' is believed, 'x*0=0' is believed, 'x*s(y)=(x*y)+x' is believed, and the rule: if "A -> B" is believed and A is believed, then (soon or later) B is believed.

`But the point of Seth Lloyd's paper was that later can effectively be never and since`

`given any time horizon, t, almost all B will not be believed earlier than t. So really you`

`call it "believe", but you use it as "provable".`

Then the theory will apply to any recursively enumerable extensions of those beliefs,provided they don't get arithmetically unsound.The believer can be shown to be Löbian once he has also the beliefs in the induction axioms.

`Not really. You have add another axiom that the believer is correct. He doesn't believe`

`any false propositions - which means it is an idealization that doesn't apply to anyone.`

Brent

Bruno

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